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# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals Exercise 7.5

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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5: Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 7 - Integrals Exercise 7.5 provide a thorough understanding of integration techniques and their applications. This exercise focuses on various methods of integrating functions, including substitution, partial fractions, and by parts. These solutions are crafted by experts to help students grasp complex concepts easily, enhancing their problem-solving skills and preparing them for board exams and competitive tests. By practicing these problems, students can build a strong foundation in calculus, ensuring a solid grasp of integrals and their properties, essential for higher studies in mathematics and related fields.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.5: Free PDF Download
2. Glance on NCERT Solutions for Class 12 Maths Ex 7.5 | Vedantu
3. Topics Covered in Maths Class 12 Ex 7.5
4. Access NCERT Solutions for Class 12 Maths Chapter 7 – Integrals Exercise 7.5
5. NCERT Solution Class 12 Maths of Chapter 7 all Exercises
6. CBSE Class 12 Maths Chapter 7 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions for Class 12 Maths Ex 7.5 | Vedantu

• Chapter 7 Exercise 7.5 of your Class 12 Maths textbook focuses on integrating rational functions using the method of partial fractions.

• Rational Functions are functions where the numerator and denominator are polynomials (expressions with variables and constant terms with whole number exponents).

• Partial Fractions breaks down a complicated rational function with a denominator of a certain form (usually a product of linear factors) into simpler fractions.

• Exercise 7.5 provides practice problems where you will use partial fractions to decompose a rational function and then integrate each of the simpler fractions. By combining the integrals of the simpler fractions, you will get the integral of the original rational function.

• Exercise 7.5 Class 12 maths NCERT Solutions has overall 23 fully solved Questions.

• Class 12 Exercise 7.5 likely involved applying these formulas to various problems.

## Topics Covered in Maths Class 12 Ex 7.5

1. Integration by Parts.

2. Integration by Partial Fractions.

1. P(x)/Q(x) dx where P(x) and Q(x) are polynomials.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 7 – Integrals Exercise 7.5

Exercise 7.5

1. Integrate $\dfrac{x}{{(x + 1)(x + 2)}}$

Ans: Let $\dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$\Rightarrow x = A(x + 2) + B(x + 1)$

Equating the coefficients of ${\text{x}}$ and constant term, we obtain $A + B = 1$

$2A + B = 0$

On solving. we obtain ${\text{A}} = - 1$ and ${\text{B}} = 2$

$\therefore \dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{{ - 1}}{{(x + 1)}} + \dfrac{2}{{(x + 2)}}$

$\Rightarrow \int {\dfrac{x}{{(x + 1)(x + 2)}}} dx = \int {\dfrac{{ - 1}}{{(x + 1)}}} + \dfrac{2}{{(x + 2)}}dx$

$= - \log |x + 1| + 2\log |x + 2| + C$

$= \log {(x + 2)^2} - \log |x + 1| + C$

$= \log \dfrac{{{{(x + 2)}^2}}}{{(x + 1)}} + C$

Where $C$ is an arbitrary constant

2. Integrate $\dfrac{1}{{{x^2} - 9}}$

Ans: Let $\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{A}{{(x + 3)}} + \dfrac{B}{{(x - 3)}}$

$1 = A(x - 3) + B(x + 3)$

Equating the coefficients of $x$ and constant term, we obtain $A + B = 0$

$- 3A + 3B = 1$

On solving. we obtain

${\text{A}} = - \dfrac{1}{6}$ and ${\text{B}} = \dfrac{1}{6}$

$\dfrac{1}{{(x + 3)(x - 3)}} = \dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}$

$\Rightarrow \int {\dfrac{1}{{\left( {{x^2} - 9} \right)}}} dx = \int {\left( {\dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}} \right)} dx$

$= - \dfrac{1}{6}\log |x + 3| + \dfrac{1}{6}\log |x - 3| + C$

$= \dfrac{1}{6}\log \dfrac{{|(x - 3)|}}{{|(x + 3)|}} + C$

Where $C$ is an arbitrary constant

3. Integrate $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}$

Ans: Let $\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$3x - 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$- 5A - 4B - 3C = 3$

$6A + 3B + 2C = - 1$

Solving these equations, we obtain ${\text{A}} = 1,\;{\text{B}} = - 5$, and $C = 4$

$\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}$

$\Rightarrow \int {\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}} \right\}} dx$

$= \log |x - 1| - 5\log |x - 2| + 4\log |x - 3| + C$

Where $C$ is an arbitrary constant.

4. Integrate $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}$

Ans: $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

$x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B + C = 0$

$45 - 3C = 1$

$6A + 4B + 2C = 0$

Solving these equations, we obtain $A = \dfrac{1}{2},B = 2$ and $C = \dfrac{3}{2}$

$\therefore \dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}$

$\Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)(x - 3)}}} dx = \int {\left\{ {\dfrac{1}{{2(x - 1)}} - \dfrac{2}{{(x - 2)}} + \dfrac{3}{{2(x - 3)}}} \right\}} dx$

$= \dfrac{1}{2}\log |x - 1| - 2\log |x - 2| + \dfrac{3}{2}\log |x - 3| + C$

Where $C$ is an arbitrary constant.

5. Integrate $\dfrac{{2x}}{{{x^2} + 3x + 2}}$

Ans: Let $\dfrac{{2x}}{{{x^2} + 3x + 2}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

$2x = A(x + 2) + B(x + 1)$

$\ldots (1)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + B = 2$

$2A + B = 0$

Solving these equations, we obtain $A = - 2$ and ${\mathbf{B}} = 4$

$\therefore \dfrac{{2x}}{{(x + 1)(x + 2)}} = \dfrac{{ - 2}}{{(x + 1)}} + \dfrac{4}{{(x + 2)}}$

$\Rightarrow \int {\dfrac{{2x}}{{(x + 1)(x + 2)}}} dx = \int {\left\{ {\dfrac{4}{{(x + 2)}} - \dfrac{2}{{(x + 1)}}} \right\}} dx$

$= 4\log |x + 2| - 2\log |x + 1| + C$

Where $C$ is an arbitrary constant.

6. Integrate $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {1 - {x^2}} \right)$ by $x(1 - 2x)$, we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{{2 - x}}{{x(1 - 2x)}}} \right)$

Let $\dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{A}{x} + \dfrac{B}{{(1 - 2x)}}$

$\Rightarrow (2 - x) = A(1 - 2x) + Bx$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $- 2A + B = - 1$

And $A = 2$ Solving these equations, we obtain $A = 2$ and $B = 3$ $\therefore \dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{2}{x} + \dfrac{3}{{1 - 2x}}$

Substituting in equation (1), we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left\{ {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right\}$

$\int {\dfrac{{1 - {x^2}}}{{x(1 - 2x)}}} dx = \int {\left\{ {\dfrac{1}{2} + \dfrac{1}{2}\left( {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right)} \right\}} dx$

$= \dfrac{x}{2} + \log |x| + \dfrac{3}{{2( - 2)}}\log |1 - 2x| + C$

$= \dfrac{x}{2} + \log |x| - \dfrac{3}{4}\log |1 - 2x| + c$

Where $C$ is an arbitrary constant.

7. Integrate $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}$

Ans: Let $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \dfrac{C}{{(x - 1)}}$

$x = (Ax + B)(x - 1) + C\left( {{x^2} + 1} \right)$

$x = A{x^2} - Ax + Bx - B + C{x^2} + C$

Equating the coefficients of ${x^2},x$, and constant term, we obtain

A $- A + B = 1$

$- B + C = 0$

On solving these equations, we obtain ${\text{A}} = - \dfrac{1}{2},\;{\text{B}} = \dfrac{1}{2}$, and ${\text{C}} = \dfrac{1}{2}$

From equation (1), vre obtain $\therefore \dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{\left( { - \dfrac{1}{2}x + \dfrac{1}{2}} \right)}}{{{x^2} + 1}} + \dfrac{{\dfrac{1}{2}}}{{(x - 1)}}$

$\Rightarrow \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}} = - \dfrac{1}{2}\int {\dfrac{x}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{{x^2} + 1}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx$

$= - \dfrac{1}{4}\int {\dfrac{{2x}}{{{x^2} + 1}}} dx + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

Consider $\int {\dfrac{{2x}}{{{x^2} + 1}}} dx$, let $\left( {{x^2} + 1} \right) = t \Rightarrow 2xdx = dt$

$\Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{{dt}}{t}} - \log |t| - \log \left| {{x^2} + 1} \right|$

$\therefore \int {\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}}} = - \dfrac{1}{4}\log \left| {{x^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}x + \dfrac{1}{2}\log |x - 1| + C$

$= \dfrac{1}{2}\log |{\text{x}} - 1| - \dfrac{1}{4}\log \left| {{{\text{x}}^2} + 1} \right| + \dfrac{1}{2}{\tan ^{ - 1}}{\text{x}} + {\text{C}}$

Where $C$ is an arbitrary constant.

8. Integrate $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}$

Ans: Let $\dfrac{x}{{{{(x - 1)}^2}(x + 2)}} - \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 2)}}$

$x = A(x - 1)(x + 2) + B(x + 2) + C{(x - 1)^2}$

Equating the coefficients of ${{\text{x}}^2},{\text{x}}$ and constant term, we obtain ${\text{A}} + {\text{C}} = 0$

$A + B - 2C = 1$

$- 2\;{\text{A}} + 2\;{\text{B}} + {\text{C}} = 0$

On solving. we obtain $A = \dfrac{2}{9}$ and $C = \dfrac{{ - 2}}{9}$

${\text{B}} = \dfrac{1}{3}$

$\therefore \dfrac{x}{{{{(x - 1)}^2}(x + 2)}} = \dfrac{2}{{9(x - 1)}} + \dfrac{1}{{3{{(x - 1)}^2}}} - \dfrac{2}{{9(x + 2)}}$

$\Rightarrow \int {\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x - 1)}}} dx + \dfrac{1}{3}\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x + 2)}}} dx$

$= \dfrac{2}{9}\log |x - 1| + \dfrac{1}{3}\left( {\dfrac{{ - 1}}{{x - 1}}} \right) - \dfrac{2}{9}\log |x + 2| + C$

$= \dfrac{2}{9}\log \left| {\dfrac{{x - 1}}{{x + 2}}} \right| - \dfrac{1}{{3(x - 1)}} + C$

Where $C$ is an arbitrary constant.

9. Integrate $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}$

Ans: $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}} = \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}$

Let $\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 1)}}$

$3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C{(x - 1)^2}$

$3x + 5 = A\left( {{x^2} - 1} \right) + B(x + 1) + C\left( {{x^2} + 1 - 2x} \right)$

Equating the coefficients of ${x^2},x$ and constant term, we obtain $A + C = 0$

$B - 2C - 3$

$- A + B + C = 5$

On solving. we obtain $B = 4$

${\text{A}} = - \dfrac{1}{2}$ and ${\text{C}} = \dfrac{1}{2}$

$\therefore \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{{ - 1}}{{2(x - 1)}} + \dfrac{4}{{{{(x - 1)}^2}}} + \dfrac{1}{{2(x + 1)}}$

$\Rightarrow \int {\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}} dx = - \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}} dx + 4\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx$

$= - \dfrac{1}{2}\log |x - 1| + 4\left( {\dfrac{{ - 1}}{{x - 1}}} \right) + \dfrac{1}{2}\log |x + 1| + C$

$= \dfrac{1}{2}\log \left| {\dfrac{{x + 1}}{{x - 1}}} \right| - \dfrac{4}{{(x - 1)}} + C$

Where $C$ is an arbitrary constant.

10. Integrate $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}$

Ans: $\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}} = \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}}$

Let $\dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{C}{{(2x + 3)}}$

$\Rightarrow (2x - 3) = A(x - 1)(2x + 3) + B(x + 1)(2x + 3) + C(x + 1)(x - 1)$

$\Rightarrow (2x - 3) = A\left( {2{x^2} + x - 3} \right) + B\left( {2{x^2} + 5x + 3} \right) + C\left( {{x^2} - 1} \right)$

$\Rightarrow (2x - 3) = (2A + 2B + C){x^2} + (A + 5B)x + ( - 3A + 3B - C)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $2A + 2B + C = 0$

$A + 5B = 2$

$- 3A + 3B - C = - 3$

On solving, we obtain ${\text{B}} = - \dfrac{1}{{10}},\;{\text{A}} = \dfrac{5}{2}$, and ${\text{C}} = - \dfrac{{24}}{5}$

$\therefore \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{5}{{2(x + 1)}} - \dfrac{1}{{10(x - 1)}} - \dfrac{{24}}{{5(2x + 3)}}$

$\Rightarrow \int {\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}} dx = \dfrac{5}{2}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{1}{{10}}\int {\dfrac{1}{{x - 1}}} dx - \dfrac{{24}}{5}\int {\dfrac{1}{{(2x + 3)}}} dx$

$= \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{24}}{{5 \times 2}}\log |2x + 3|$

$= \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{12}}{5}\log |2x + 3| + C$

Where $C$ is an arbitrary constant.

11. Integrate $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}$

Ans : $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}} = \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}}$

Let $\dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}} + \dfrac{C}{{(x - 2)}}$

$5x = A(x + 2)(x - 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $A + B + C = 0$

$B + 3C = 5$ and $4A - 2B + 2C = 0$

On solving. we obtain $A - \dfrac{5}{3},B - - \dfrac{5}{2}$, and $C - \dfrac{5}{6}$

$\therefore \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{5}{{3(x + 1)}} + - \dfrac{5}{{2(x + 2)}} + \dfrac{5}{{6(x - 2)}}$

$\Rightarrow \int {\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}} dx = \dfrac{5}{3}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{5}{2}\int {\dfrac{1}{{(x + 2)}}} dx + \dfrac{5}{6}\int {\dfrac{1}{{(x - 2)}}} dx$

$= \dfrac{5}{3}\log |x + 1| - \dfrac{5}{2}\log |x + 2| + \dfrac{5}{6}\log |x - 2| + C$

Where $C$ is an arbitrary constant.

12. Integrate $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}}$

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {{x^3} + x + 1} \right)$ by ${x^2} - 1$, we obtain $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{{2x + 1}}{{{x^2} - 1}}$

Let $\dfrac{{2x + 1}}{{{x^2} - 1}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}}$

$2x + 1 = A(x - 1) + B(x + 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 2$

$- A + B = 1$

On solving. we obtain $A - \dfrac{1}{2}$ and $B - \dfrac{3}{2}$

$\therefore \dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{1}{{2(x + 1)}} + \dfrac{3}{{2(x - 1)}}$

$\Rightarrow \int {\dfrac{{{x^3} + x + 1}}{{{x^2} + 1}}} dx = \int x dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx + \dfrac{3}{2}\int {\dfrac{1}{{(x - 1)}}} dx$

$= \dfrac{{{x^2}}}{2} + \log |x + 1| + \dfrac{3}{2}\log |x - 1| + C$

Where $C$ is an arbitrary constant.

13. Integrate $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}$

Ans:

Let $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(1 - x)}} + \dfrac{{Bx + C}}{{\left( {1 + {x^2}} \right)}}$

$2 = A\left( {1 + {x^2}} \right) + (Bx + C)(1 - x)$

$2 = A + A{x^2} + Bx - B{x^2} + C - Cx$

Equating the coefficient of ${x^2},x$, and constant term, we obtain ${\text{A}} - {\text{B}} = 0$

${\mathbf{B}} - {\mathbf{C}} = {\mathbf{0}}$

$A + C = 2$

On solving these equations, we obtain $A = 1,B = 1$, and $C = 1$

$\therefore \dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{1}{{1 - x}} + \dfrac{{x + 1}}{{1 + {x^2}}}$

$\Rightarrow \int {\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}} dx = \int {\dfrac{1}{{1 - x}}} dx + \int {\dfrac{x}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$= - \int {\dfrac{1}{{1 - x}}} dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$= - \log |x - 1| + \dfrac{1}{2}\log \left| {1 + {x^2}} \right| + {\tan ^{ - 1}}x + C$

Where $C$ is an arbitrary constant.

14. Integrate $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}$

Ans:

Let $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{{{(x + 2)}^2}}}$

$\Rightarrow 3x - 1 = A(x + 2) + B$

Equating the coefficient of $x$ and constant term, we obtain $A = 3$

$2A + B = - 1 \Rightarrow \mathbb{B} - - 7$

$\therefore \dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{3}{{(x + 2)}} - \dfrac{7}{{{{(x + 2)}^2}}}$

$\Rightarrow \int {\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}} dx = 3\int {\dfrac{1}{{(x + 2)}}} dx - 7\int {\dfrac{1}{{{{(x + 2)}^2}}}} dx$

$= 3\log |x + 2| - 7\left( {\dfrac{{ - 1}}{{(x + 2)}}} \right) + C$

$= 3\log |x + 2| + \dfrac{7}{{(x + 2)}} + C$

Where $C$ is an arbitrary constant.

15. Integrate $\dfrac{1}{{{x^4} - 1}}$

Ans :

$\dfrac{1}{{\left( {{x^4} - 1} \right)}} - \dfrac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} - \dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}}$

Let $\dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 1} \right)}}$

$1 = A(x - 1)\left( {1 + {x^2}} \right) + B(x + 1)\left( {1 + {x^2}} \right) + (Cx + D)\left( {{x^2} - 1} \right)$

$1 = A\left( {{x^3} + x - {x^2} - 1} \right) + B\left( {{x^3} + x + {x^2} + 1} \right) + C{x^3} + D{x^2} - Cx - D$

$1 = (A + B + C){x^3} + ( - A + B + D){x^2} + (A + B - C)x + ( - A + B - D)$

Equating the coefficient of ${x^3},{x^2},x$, and constant term, we obtain $A + B + C = 0$

$- A + B + D = 0$

$A + B - C = 0$

$- A + B - D = 1$

${\text{A}} = - \dfrac{1}{4},\;{\text{B}} = \dfrac{1}{4},{\text{C}} = {\text{O}}$, and ${\text{D}} = - \dfrac{1}{2}$

$\therefore \dfrac{1}{{\left( {{x^n} - 1} \right)}} = \dfrac{{ - 1}}{{4(x + 1)}} + \dfrac{1}{{4(x - 1)}} + \dfrac{1}{{2\left( {{x^2} + 1} \right)}}$

$\Rightarrow \int {\dfrac{1}{{{x^4} - 1}}} dx - - \dfrac{1}{4}\log |x - 1| + \dfrac{1}{4}\log |x - 1| - \dfrac{1}{2}{\tan ^1}x + C$

$= \dfrac{1}{4}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| - \dfrac{1}{2}{\tan ^1}x + C$

Where $C$ is an arbitrary constant.

16. Integrate $\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Hint: multiply numerator and denominator by ${x^{n - 1}}$ and put $x^n = t$

Ans:

$\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Multiplying numerator and denominator by ${x^{n - 1}}$, we obtain $\dfrac{1}{{x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^{n - 1}}x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}$

Let ${x^n} = t \Rightarrow n{x^{ - 1}}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}} dx = \int {\dfrac{{{x^{n - 1}}}}{{{x^7}\left( {{x^n} + 1} \right)}}} dx = \dfrac{1}{n}\int {\dfrac{1}{{t(t + 1)}}} dt$

Let $\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t + 1)}}$

$1 = A(1 + t) + Bt$

Equating the coefficients of $t$ and constant, ve obtain $A = 1$ and $B = - 1$

$\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{(1 + t)}}$

$\Rightarrow \int {\dfrac{1}{{x\left( {{x^4} + 1} \right)}}} dx = \dfrac{1}{n}\int {\left\{ {\dfrac{1}{t} - \dfrac{1}{{(1 + t)}}} \right\}} dx$

$\dfrac{1}{n}[\log |t| - \log |t + 1|] + C$

$= \dfrac{1}{n}\left[ {\log \left| {{x^n}} \right| - \log \left| {{x^n} + 1} \right|} \right] + C$

$= \dfrac{1}{n}\log \left| {\dfrac{{{x^n}}}{{{x^2} + 1}}} \right| + C$

Where $C$ is an arbitrary constant.

17. Integrate $\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

(Hint: Put $\sin x = t]$

Ans:

$\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

let $\sin x = t \Rightarrow \cos xdx = dt$

$\therefore \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}}$

Let $\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}$

$1 = A(2 - t) + B(1 - t)$

Equating the coefficients of $t$ and constant, wre obtain $- A - B = 0$ and $2A + B = 1$

On solving. we obtain $A = 1$ and $B = - 1$

$\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}$

$\Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\left\{ {\dfrac{1}{{1 - t}} - \dfrac{1}{{(2 - t)}}} \right\}} dt$

$= \log \left| {\dfrac{{2 - t}}{{1 - t}}} \right| + C$

$= \log \left| {\dfrac{{2 - \sin x}}{{1 - \sin x}}} \right| + C$

Where $C$ is an arbitrary constant.

18. Integrate $\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Ans:

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \dfrac{{4{x^2} + 10}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Let $\dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 3} \right)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}$

$4{x^2} + 10 = (Ax + B)\left( {{x^2} + 4} \right) + (Cx + D)\left( {{x^2} + 3} \right)$

$4{x^2} + 10 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + 3Cx + D{x^2} + 3D$

$4{x^2} + 10 = (A + C){x^2} + (B + D){x^2} + (4A + 3C)x + (4B + 3D)$

Equating the coefficients of ${x^3},{x^2},x$ and constant term, we obtain $A + C = 0$

$B + D = 4$

$4A + 3C = 0$

$4B + 3D = 10$

On solving these equations, we obtain ${\text{A}} = 0,\;{\text{B}} = - 2,{\text{C}} = 0$, and ${\text{D}} = 6$

$\therefore \dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}$

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \left( {\dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right)$

$\Rightarrow \int {\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}} dx = \int {\left\{ {1 + \dfrac{2}{{\left( {{x^2} + 3} \right)}} - \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right\}} dx$

$= \int {\left\{ {1 + \dfrac{2}{{{x^2} + {{(\sqrt 3 )}^2}}} - \dfrac{6}{{{x^2} + {2^2}}}} \right\}}$

$= x + 2\left( {\dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{x}{{\sqrt 3 }}} \right) - 6\left( {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{x}{2}} \right) + C$

$= x + \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\dfrac{x}{2} + C$

Where $C$ is an arbitrary constant.

19. Integrate $\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Ans:

$\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Let ${x^2} = t \Rightarrow 2xdx = dt$

$\therefore \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\dfrac{{dt}}{{(t + 1)(t + 3)}}}$

Let $\dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{A}{{(t + 1)}} + \dfrac{B}{{(t + 3)}}$

$1 = A(t + 3) + B(t + 1)$

Equating the coefficients of ${\text{t}}$ and constant, we obtain $A + B = 0$ and $3A + B = 1$

On solving. we obtain ${\text{A}} = \dfrac{1}{2}$ and ${\text{B}} = - \dfrac{1}{2}$

$\therefore \dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}$

$\Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\left\{ {\dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}} \right\}} dt$

$= \dfrac{1}{2}\log |(t + 1)| - \dfrac{1}{2}\log |t + 3| + C$

$= \dfrac{1}{2}\log \left| {\dfrac{{t + 1}}{{t + 3}}} \right| + C$

$= \dfrac{1}{2}\log \left| {\dfrac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C$

Where $C$ is an arbitrary constant.

20. Integrate $\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Ans:

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Multiplying numerator and denominator by ${x^3}$, we obtain

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}} = \dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \int {\dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}} dx$

Let ${x^4} = t \Rightarrow 4{x^3}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\dfrac{{dt}}{{t(t - 1)}}}$

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t - 1)}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of $t$ and constant, we obtain $A + B = 0$ and $- A = 1$

${\text{A}} = - 1$ and ${\mathbf{B}} = 1$

$\Rightarrow \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$\Rightarrow \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\left\{ {\dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}} \right\}} dt$

$- \dfrac{1}{4}[ - \log |t| + \log |t - 1|] + C$

$= \dfrac{1}{4}\log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$= \dfrac{1}{4}\log \left| {\dfrac{{{x^4} - 1\mid }}{{{x^4}}}} \right| + C$

Where $C$ is an arbitrary constant.

21. Integrate $\dfrac{1}{{\left( {{{\text{e}}^ x } - 1} \right)}}$

Hint: Put $\dfrac{1}{e^x - 1}$

Ans :

Let ${e^x} = t \Rightarrow {e^x}dx = dt$

$\Rightarrow \int {\dfrac{1}{{\left( {{{\text{e}}^x} - 1} \right)}}} {\text{dx}} = \int {\dfrac{1}{{{\text{t}} - 1}}} \times \dfrac{{{\text{dt}}}}{{\text{t}}} - \int {\dfrac{1}{{{\text{t}}({\text{t}} - 1)}}} {\text{dt}}$

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{t - 1}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of ${\text{t}}$ and constant, we obtain

$A + B = 0$ and $- A = 1$

$A = - 1$ and ${\mathbf{B}} = {\mathbf{1}}$

$\therefore \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$\Rightarrow \int {\dfrac{1}{{t(t - 1)}}} dt = \log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$= \log \left| {\dfrac{{{{\text{e}}^x} - 1}}{{{{\text{e}}^x}}}} \right| + {\text{C}}$

Where $C$ is an arbitrary constant.

22. $\int {\dfrac{{xdx}}{{(x - 1)(x - 2)}}}$ equals

a. $A\log \left| {\dfrac{{{{(x - 1)}^2}}}{{x - 2}}} \right| + C$

b. $\log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

c. $\log \left| {{{\left( {\dfrac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$

d. $\log |(x - 1)(x - 2)| + C$

Ans:

Let $\dfrac{x}{{(x - 1)(x - 2)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}}$

$x = A(x - 2) + B(x - 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 1$ and $- 2A - B = 0$

$A--1$ and $B = 2$

$\dfrac{x}{{(x - 1)(x - 2)}} = - \dfrac{1}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}$

$\Rightarrow \int {\dfrac{x}{{(x - 1)(x - 2)}}} dx = \int {\left\{ {\dfrac{{ - 1}}{{(x - 1)}} + \dfrac{2}{{(x - 2)}}} \right\}} dx$

$= - \log |x - 1| + 2\log |x - 2| + C$

$= \log \left| {\dfrac{{{{(x - 2)}^2}}}{{x - 1}}} \right| + C$

Hence, the correct Answer is $B$.

23. $\int {\dfrac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$ equals

a. $\log |x| - \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

b. $\log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

c. $- \log |x| + \dfrac{1}{2}\log \left( {{x^2} + 1} \right) + C$

d. $\dfrac{1}{2}\log |x| + \log \left( {{x^2} + 1} \right) + C$

Ans :

Let $\dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{A}{x} + \dfrac{{Bx + C}}{{{x^2} + 1}}$

$1 = A\left( {{x^2} + 1} \right) + (Bx + C)x$

Equating the coefficients of ${x^2},x$, and constant term, we obtain $A + B = 0$

$c = 0$

$A = 1$

On solving these equations, we obtain $A = 1,B = - 1$, and $C = 0$

$\therefore \dfrac{1}{{x\left( {{x^2} + 1} \right)}} = \dfrac{1}{x} + \dfrac{{ - x}}{{{x^2} + 1}}$

$\Rightarrow \int {\dfrac{1}{{x\left( {{x^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{1}{x} - \dfrac{x}{{{x^2} + 1}}} \right\}} dx$

$- \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Hence, the correct Answer is ${\text{A}}$.

Where $C$ is an arbitrary constant.

Alter:

$\Rightarrow \int {\dfrac{1}{{{\text{x}}\left( {{{\text{x}}^2} + 1} \right)}}} dx = \int {\left\{ {\dfrac{{\text{x}}}{{{{\text{x}}^2}\left( {{{\text{x}}^2} + 1} \right)}}} \right\}} {\text{dx}}$

Let ${x^2} = t$, therefore, $2xdx = dt$ $\therefore \int {\dfrac{x}{{{x^2}\left( {{x^2} + 1} \right)}}} dx = \dfrac{1}{2}\int {\dfrac{{dt}}{{t(t + 1)}}} = \dfrac{1}{2}\int {\dfrac{{(t + 1) - t}}{{t(t + 1)}}} dt = \dfrac{1}{2}\int \frac{1}{t} - \dfrac{1}{{t + 1}}dt$

$- \dfrac{1}{2}[\log t - \log (t + 1)] + C$

$= \log |x| - \dfrac{1}{2}\log \left| {{x^2} + 1} \right| + C$

Where $C$ is an arbitrary constant.

## Conclusion

The NCERT Solutions for Class 12 Chapter 7 Integrals Exercise 7.5 focuses on the application of integration techniques to solve various types of integral problems, particularly focusing on integration by parts and the integration of rational functions. It's essential to understand the fundamental integration formulas and the steps for applying integration by parts, as these are commonly used methods in this exercise. Emphasizing practice on these techniques will help in solving the problems efficiently.

## NCERT Solution Class 12 Maths of Chapter 7 all Exercises

 Exercises Number of Questions Exercise 7.1 22 Questions & Solutions (21 Short Answers, 1 MCQs) Exercise 7.2 39 Questions & Solutions (37 Short Answers, 2 MCQs) Exercise 7.3 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.4 25 Questions & Solutions (23 Short Answers, 2 MCQs) Exercise 7.6 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.7 11 Questions & Solutions (9 Short Answers, 2 MCQs) Exercise 7.8 22 Questions & Solutions (20 Short Answers, 2 MCQs) Exercise 7.9 10 Questions & Solutions (8 Short Answers, 2 MCQs) Exercise 7.10 21 Questions & Solutions (19 Short Answers, 2 MCQs)

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 7 - Integrals Exercise 7.5

1. What are the two types of Integrals?

The two types of Integrals are

• Definite Integrals: An integral that is to be evaluated between a lower limit a and upper limit b.

For example, $\int_{a}^{b} f\left ( x \right )dx$

• Indefinite Integrals: An integral which does not have any upper and lower limit.

For example, $\int f\left ( x \right )dx$

2. What are the properties of definite integrals that are involved in Exercise 7.5?

Some key properties include:

Linearity:

$\int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx$

Reversal of limits:

$\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx$

$\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx$

Symmetry: For even functions $$f(x)$$,

$\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$

3. What is the difference between definite and indefinite integrals?

An indefinite integral represents a family of functions and includes a constant of integration (C), like ∫f(x) dx=F(x)+C.

A definite integral calculates the exact area under a curve between two points and does not include C, like$\int_{a}^{b}f\left ( x \right )dx$.

4. What is the main focus of Exercise 7.5 in Chapter 7 on Integrals?

Exercise 7.5 focuses on evaluating definite integrals using the properties of definite integrals. This means you'll be solving integrals within specific limits (from a to b) and applying various properties to simplify and solve them.

5. How to solve definite integrals involving trigonometric functions?

Step 1: SImplify the integrand by using trigonometric identities like

$\sin^2 x = \frac{1 - \cos 2x}{2}$

$\cos^2 x = \frac{1 + \cos 2x}{2}$

Step 2: Sometimes, using substitution (like u=sin⁡x or u=cos⁡x) can make the integral easier to solve.

Step 3:  Utilize symmetry properties for integrals involving trigonometric functions over symmetric limits.